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          虎年快乐之算法练习题22---动态规划“最少硬币问题”
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        <h1 id="前言"><a href="#前言" class="headerlink" title="前言"></a>前言</h1><p>2022虎年初一，祝大家新年快乐！今天借着喜气，写一篇关于算法竞赛中非常重要的一种思想—-动态规划（DP）。它主要用来解决最优解问题，而其思想核心往往是把最优解细化成许多个子问题来求解，最后在一步步的利用前面的结果递进优化得到问题的最优解。而这些子问题是前后相关的，并且非常相似，且处理方法一样。下面通过非常经典的一道例题来介绍这种算法思想。</p>
<span id="more"></span>

<h1 id="一、题目描述"><a href="#一、题目描述" class="headerlink" title="一、题目描述"></a>一、题目描述</h1><p>有n种硬币，面值分别为v1，v2，…，vn，数量无限。输入非负整数s，选用硬币，使其和为s。要求输出最少的硬币组合。</p>
<h4 id="输入样例："><a href="#输入样例：" class="headerlink" title="输入样例："></a>输入样例：</h4><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">5</span><br><span class="line">1 5 10 25 50</span><br><span class="line">251</span><br></pre></td></tr></table></figure>

<h4 id="输出样例："><a href="#输出样例：" class="headerlink" title="输出样例："></a>输出样例：</h4><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">6</span><br></pre></td></tr></table></figure>

<h1 id="二、DP思路"><a href="#二、DP思路" class="headerlink" title="二、DP思路"></a>二、DP思路</h1><p>我们可以定义一个数组，int Min[MONEY]，其中Min[i]对应着在凑够金额 i 下需要的最少硬币个数是多少，有了这个数组，我们就可以用来记录每一个数值下，硬币个数的最优解。动规一开始要进行初始化操作，思路就是先尝试用面值最小的硬币去装填Min[i]，这样得出来的解一定是硬币最多的情况。</p>
<p>例如：现在有（1,5,10）三种面值的硬币</p>
<p>那么首先用1元面值去装填，那么Min[i]等于i（Min[0]=0），这样一定不是最优解，那么把所有金额装完后，开始尝试用5元面值去装，那么在Min[1]~Min[4]仍然用1元是最优解，但是在Min[5]时，Min[5]=min（Min[5-5]+1,Min[5]）,那么Min[5]的结果显然会变为1，刚才我所用的这样一个取最小的判断过程，就是动态规划中非常重要的<strong>状态转移方程</strong>。</p>
<p>那么依次类推，同样是在5元面值下，Min[6]=Min[6-5]+1=2，这样最终我们就可以得到所有金额的最优解个数</p>
<h1 id="三、具体代码"><a href="#三、具体代码" class="headerlink" title="三、具体代码"></a>三、具体代码</h1><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span><span class="meta-string">&lt;bits/stdc++.h&gt;</span></span></span><br><span class="line"><span class="keyword">using</span> <span class="keyword">namespace</span> <span class="built_in">std</span>;</span><br><span class="line"><span class="keyword">int</span> n;</span><br><span class="line"><span class="keyword">int</span> money;</span><br><span class="line"><span class="keyword">int</span> Min[<span class="number">1000000</span>]; <span class="comment">//每个金额对应最少的硬币数量</span></span><br><span class="line"><span class="keyword">int</span> value[<span class="number">10</span>];  <span class="comment">//定义面值数组</span></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">solve</span><span class="params">()</span></span></span><br><span class="line"><span class="function"></span>&#123;</span><br><span class="line">	<span class="keyword">for</span>(<span class="keyword">int</span> k=<span class="number">0</span>;k&lt;=money;k++)  <span class="comment">//初始值设为无穷大</span></span><br><span class="line">	&#123;</span><br><span class="line">		Min[k]=INT_MAX;</span><br><span class="line">	&#125;</span><br><span class="line">	Min[<span class="number">0</span>]=<span class="number">0</span>;</span><br><span class="line">	<span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j&lt;=n;j++)</span><br><span class="line">	&#123;</span><br><span class="line">		<span class="keyword">for</span>(<span class="keyword">int</span> i=value[j];i&lt;=money;i++)</span><br><span class="line">		&#123;</span><br><span class="line">			Min[i]=min(Min[i],Min[i-value[j]]+<span class="number">1</span>);  <span class="comment">//比较使用value[j]这种面值的硬币后，需要的硬币数量是否减少</span></span><br><span class="line">		&#125;</span><br><span class="line">	&#125;</span><br><span class="line">&#125;</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span></span><br><span class="line"><span class="function"></span>&#123;</span><br><span class="line">	<span class="built_in">cin</span>&gt;&gt;n; <span class="comment">//n种硬币</span></span><br><span class="line">	<span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i&lt;=n;i++)</span><br><span class="line">	&#123;</span><br><span class="line">		<span class="built_in">cin</span>&gt;&gt;value[i];</span><br><span class="line">	&#125;</span><br><span class="line">	<span class="built_in">cin</span>&gt;&gt;money;</span><br><span class="line">	solve();</span><br><span class="line">	<span class="built_in">cout</span>&lt;&lt;Min[money]&lt;&lt;<span class="built_in">endl</span>;</span><br><span class="line">	<span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>


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